# Homework Help: Maximum Value of gradient

1. Nov 18, 2006

### team31

when u=-0.6i+0.8j, Du f(1,2)=? gradient f=(18,9)
I use unit vector dot (-0.6,0.8) dotted with the gradient f, which turns out to be -3.6
so here is the question. A: the maximum value of Dv f(1,2) is=?
B: this maximum occurs when v=(?,?)
I'm confued with the Dv thing. For my guess is that I should use (1,2) to creat a unit vector which is (1/sqrt5, 2/sqrt5), and use that unit vetor dotted with the gradient f, which gives me the maximum value of Dv f(1,2)
then i will find a vector that its dot product with ( 1/sqrt5, 2/sqrt5) is 0, and tha should be the direction of V which is the answer for part B

Am I doing this right? plaese help me out. I've been looking into my text book but couldn't find the symbol Dv, they all used Du. Thank u

2. Nov 21, 2006

### benorin

Directional Derivatives

Directional Derivatives

Directional derivatives of $f(x,y)$ are, for unit vectors $$\vec{u}$$, are defined by:

$$D_{\vec{u}}f(x,y) = \vec{u}\cdot \vec{\nabla}f (x,y)$$​

where $$\cdot$$ denotes the dot product. To see when the maximum value of the directional derivatives occurs, consider the magnitude of the directional derivative, recalling that $$\left|\vec{u}\cdot\vec{v} \right| = \left|\vec{u}\right|\cdot \left|\vec{v} \right|\cos \theta$$ where $$\theta$$ is the acute angle between the vectors we have that

$$\left| D_{\vec{u}}f(x,y)\right| = \left| \vec{u}\cdot \vec{\nabla}f (x,y)\right| = \left| \vec{u}\right|\cdot\left| \vec{\nabla}f (x,y)\right|\cos \theta = 1\cdot\left| \vec{\nabla}f (x,y)\right|\cos \theta = \left| \vec{\nabla}f (x,y)\right|\cos \theta$$​

since $$\left| \vec{u}\right| =1$$ since $$\vec{u}$$ is a unit vector; consider that the quantity $$\left| \vec{\nabla}f (x,y)\right|\cos \theta$$ is maximized when $$\cos \theta =1$$, i.e. when $$\theta =0$$ namely when the vectors $$\vec{u}$$ and $$\vec{\nabla}f (x,y)$$ are paralell. This occurs if $$\vec{u} = \frac{\vec{\nabla}f (x,y)}{\left| \vec{\nabla}f (x,y)\right|}$$ which is a unit vector paralell to $$\vec{\nabla}f (x,y)$$. The maximum value is then

$$D_{\frac{\vec{\nabla}f (x,y)}{\left| \vec{\nabla}f (x,y)\right|}}f(x,y) = \frac{\vec{\nabla}f (x,y)}{\left| \vec{\nabla}f (x,y)\right|}\cdot \vec{\nabla}f (x,y) = \frac{\left| \vec{\nabla}f (x,y)\right|^{2}}{\left| \vec{\nabla}f (x,y)\right|} = \left| \vec{\nabla}f (x,y)\right|$$​

Last edited: Nov 21, 2006